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A massless four-momentum is not obtained from the massive formula at m = 0

Statement refuted

The four-momentum of an ideal massless particle is obtained by setting $m = 0$ in $P = m U$ , where $U$ is a future timelike four-velocity with $η (U, U) = - c^{2}$ .

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ ; a future timelike $U$ with $η (U, U) = - c^{2}$ (Four-velocity and four-acceleration of a future timelike worldline); the observer quantities $E_{n}$ of Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum for a future unit observer $n$ .

Here massless four-momentum means a nonzero future null vector of $V$ ; the refutation is purely algebraic and asserts no physical interpretation.

[F1]

(Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, Definition): $η$ is a symmetric bilinear form on $V$ , so $η (v, 0) = 0$ and $η (0, 0) = 0$ for all $v$ , and $η (λ v, μ w) = λ μ η (v, w)$ for real $λ, μ$ ; in an $η$ -orthonormal ordered basis $η (e_{0}, e_{0}) = - 1$ , $η (e_{1}, e_{1}) = 1$ and $η (e_{0}, e_{1}) = 0$ .

[F2]

(Timelike, null, spacelike and causal vectors of a Lorentz vector space, Definition): $v$ is timelike when $η (v, v) < 0$ , null when $v \neq 0$ and $η (v, v) = 0$ , spacelike when $η (v, v) > 0$ , and the zero vector is none of timelike, null or spacelike.

[F3]

(Inertial frames and inertial charts of the affine Minkowski model, Definition): an inertial frame's basis $(e_{0}, e_{1}, e_{2}, e_{3})$ is $η$ -orthonormal with $e_{0} \in C^{+}$ , so $e_{0}$ is future timelike and $η (e_{0}, e_{0}) = - 1$ .

[F4]

(The timelike set has exactly two convex components, described in orthonormal coordinates, Statement): in an $η$ -orthonormal basis whose first vector is timelike, $c l (C^{+}) = {v : v^{0} \geq ∣ v_{s} ∣}$ for the naming of $C^{+}$ containing the first vector, where $v = v^{0} e_{0} + v_{s}$ and $∣ v_{s} ∣ = \sqrt{(v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}}$ .

[F5]

(Time orientation, future and past timelike/null/causal vectors, and the closed cones, Definition): a vector is future null when it is nonzero, $η (v, v) = 0$ and $v$ lies in $c l (C^{+})$ , and future causal when it is future timelike or future null.

[F6]

(Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum, Definition): $E_{n} (P) = - c η (P, n)$ for $P \in c l (C^{+})$ .

[F7]

(Positivity, mass shell and kinetic energy of the observer quantities of a four-momentum, Statement (a)): for nonzero future causal $P$ , $E_{n} (P) > 0$ and, for future null $P$ , $E_{n} (P) = c ∣ q_{n} (P) ∣$ .

Counterexample

1.1F1F2F6

The case $m = 0$ : for the future timelike $U$ one has $0 \cdot U = 0$ , so the massive formula at $m = 0$ produces the zero vector; by [F1] and [F6], $η (0, 0) = 0$ and $E_{n} (0) = - c η (0, n) = 0$ , and by [F2] the zero vector is not a nonzero future null vector since it is not null (nullity requires $P \neq 0$ ).

1.2F1F3F4F5F6F7

The witness of a genuine massless momentum: in the inertial frame of [F3] choose $E > 0$ and set $P : = (E / c) (e_{0} + e_{1})$ . Then [F1] gives $η (P, P) = (E / c)^{2} (η (e_{0}, e_{0}) + 2 η (e_{0}, e_{1}) + η (e_{1}, e_{1})) = (E / c)^{2} (- 1 + 0 + 1) = 0$ and $P \neq 0$ ; its coordinates are $(v^{0}, v^{1}, v^{2}, v^{3}) = (E / c) (1, 1, 0, 0)$ , so $v^{0} = E / c = ∣ v_{s} ∣$ and [F4] places $P$ in $c l (C^{+})$ ; hence $P$ is future null by [F5]. Moreover $E_{e_{0}} (P) = - c η (P, e_{0}) = - (E) (η (e_{0}, e_{0}) + η (e_{1}, e_{0})) = E > 0$ by [F1] and [F6], in agreement with [F7].

2.1F1F2step 1.1step 1.2

The case $m > 0$ : for every $m > 0$ , [F1] gives $η (m U, m U) = m^{2} η (U, U) = - m^{2} c^{2} < 0$ , so $m U$ is timelike by [F2]; if $m U = P$ then $η (m U, m U) = η (P, P) = 0$ , a contradiction with $m U$ timelike. Therefore no $m > 0$ produces the future null witness $P = m U$ , and by [step 1.1] no $m = 0$ does either: no value $m \geq 0$ in the massive formula $P = m U$ produces the massless momentum $P$ .

3.1step 1.1step 1.2step 2.1∎

Combining the two cases, the statement is refuted: setting $m = 0$ in $P = m U$ gives the zero vector, which is not a nonzero future null four-momentum and has zero energy in every frame, while a nonzero future null four-momentum such as the witness of [step 1.2] is not of the form $m U$ for any $m \geq 0$ and any future timelike $U$ with $η (U, U) = - c^{2}$ . This refutes direct substitution with a defined normalized timelike $U$ ; the argument leaves mass-dependent limits and physical formulation choices open.

A massless four-momentum is not obtained from the massive formula at m = 0

Statement refuted

Facts & Assumptions

Counterexample

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